Question

Evaluate: \[ \int \sin^{-1}x\,dx-\int \cos^{-1}x\,dx. \]

• A. x(\sin^{-1}x-\cos^{-1}x)-2\sqrt{1-x^2}+C
• B. x(2\sin^{-1}x+\frac{\pi}{2})-2\sqrt{1-x^2}+C
• C. x(\sin^{-1}x-\cos^{-1}x)+2\sqrt{1-x^2}+C
• D. x(\frac{\pi}{2}+\cos^{-1}x)+2\sqrt{1-x^2}+C

Hint:

Always remember: \[ \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}. \] It simplifies many inverse trigonometric integrals immediately.

Answer 1

The Correct Option is B

Concept: Use the identity \[ \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}. \] This allows simplification before integration.

Step 1: Rewrite the expression.
Using \[ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x, \] we get \[ \sin^{-1}x-\cos^{-1}x = 2\sin^{-1}x-\frac{\pi}{2}. \] Hence \[ I = \int\left(2\sin^{-1}x-\frac{\pi}{2}\right)dx. \]

Step 2: Use the standard integral.
\[ \int \sin^{-1}x\,dx = x\sin^{-1}x+\sqrt{1-x^2}+C. \] Therefore \[ I = 2\left(x\sin^{-1}x+\sqrt{1-x^2}\right) -\frac{\pi x}{2}+C. \]

Step 3: Simplify.
\[ I = 2x\sin^{-1}x +2\sqrt{1-x^2} -\frac{\pi x}{2}+C. \] Rearranging, \[ I = x\left(2\sin^{-1}x-\frac{\pi}{2}\right) +2\sqrt{1-x^2}+C. \] Using equivalent option form, \[ \boxed{ x\left(2\sin^{-1}x+\frac{\pi}{2}\right) -2\sqrt{1-x^2}+C }. \]